With -s switch, then any switch to the script will be treated as boolean switch unless supplied with argument. So -foo will set $foo to 1, and -foo=anything will set $foo to anything. 
$ cat foo.pl
#!/usr/bin/perl -s
use strict;
use vars '$foo';
print $foo, "\n";

$ ./foo.pl -foo
1
$ ./foo.pl -foo=bar
bar

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In reply to Re: unexpected results using -s switch by naikonta
in thread unexpected results using -s switch by mojodaddy

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